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4 years ago

Light is traveling in air at a speed of 300,000 km/s. Which substance would cause the light to refract the LEAST?

1 answer:

6 0

Use the formula n = c / v, where c is the speed of light and v is the speed of light in the environment, that is, use the speed in brackets and the correct answer will be the one that will be the smallest result

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An inductor with an inductance of 2.30H and a resistance of 8.00 Ω isconnected to the terminals of a battery with an emf of 6.00

Basile [38]

Answer:

(a). The initial rate is 2.60 A/s.

(b). The rate of current increases is 0.8658 A/s.

(c). The current is 0.435 A.

(d). The final steady-state current is 0.75 A.

Explanation:

Given that,

Inductance = 2.30 H

Resistance = 8.00 Ω

Voltage = 6.00 V

(a). We need to calculate the initial rate of increase of current in the circuit.

Using formula of initial rate

$V=initial\ rate\times inductance$

$initial\ rate=\dfrac{V}{L}$

Put the value into the formula

$initial \rate=\dfrac{6.00}{2.30}$

$initial\ rate=2.60\ A/s$

The initial rate is 2.60 A/s.

(b). We need to calculate the rate of increase of current at the instant when the current is 0.500

Using formula of rate of increase of current

$rate\ of \ current\ increase=initial\ rate\times e^{\dfrac{-t}{T}}$ ....(I)

Where, $T=\dfrac{L}{R}$

$T=\dfrac{2.30}{8.00}$

$T=0.2875$

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

$e^{\dfrac{-t}{T}}=1-(i\times\dfrac{R}{V})$

$e^{\dfrac{-t}{T}}=1-(0.500\times\dfrac{8.00}{6.00})$

$e^{\dfrac{-t}{T}}=0.333$

The rate of current increases is

Put the value in the equation (I)

$rate\ of\ current\ increase=2.60\times0.333$

$rate\ of\ current\ increase=0.8658\ A/s$

The rate of current increases is 0.8658 A/s.

(c). We need to calculate the current

Using formula of current

$i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})$

Put the value into the formula

$i=\dfrac{6.00}{8.00}\times(1-e^{\dfrac{-0.250}{0.2875}})$

$i=0.435\ A$

The current is 0.435 A.

(d). We need to calculate the final steady-state current

Using formula of steady state

$i=\dfrac{V}{R}$

$i=\dfrac{6.00}{8.00}$

$i=0.75\ A$

The final steady-state current is 0.75 A.

Hence, This is the required solution.

3 0

3 years ago

One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R

Tanzania [10]

(a) $24.6 m/s^2$

At a distance r=R from the centre of the planet, there is no effect due to the outer shell: so, the gravitational field strength at r=R is only determined by the gravity produced by the core of the planet.

So, the strength of the gravitational field is given by

$g= \frac{GM}{R^2}$

where

G is the gravitational constant

M = 6.24 × 10^24 kg is the mass of the core of the planet

R = 4.11 × 10^6 m is the radius of the core

Substituting into the equation, we find

$g= \frac{(6.67\cdot 10^{-11})(6.24\cdot 10^{24} kg)}{(4.11\cdot 10^6 m)^2}=24.6 m/s^2$

(b) $13.7 m/s^2$

at distance r=3R from the centre, the particle feels the effect of gravity due to both the core of the planet and the outer shell between R and 2R.

So, we have to consider the total mass that exerts the gravitational attraction at r=3R, which is the sum of the mass of the core (M) and the mass of the shell (4M):

M' = M + 4M = 5M

Therefore, the gravitational acceleration at r=3R will be

$g'= \frac{G(5M)}{(3R)^2}=\frac{5}{9}\frac{GM}{R^2} = \frac{5}{9}g$

And susbstituting

g = 24.6 m/s^2

found in the previous part, we find

$g' = \frac{5}{9} (24.6 m/s^2)=13.7 m/s^2$

6 0

3 years ago

What advantage does absolute-age dating have over relative-age dating? Explain

kramer

Relative age is the age of a rock layer (or the fossils it contains) compared to other layers. It can be determined by looking at the position of rock layers. Absolute age is the numeric age of a layer of rocks or fossils. Absolute age can be determined by using radiometric dating.

8 0

3 years ago

Cations are formed by gaining protons losing protons gaining electrons losing electrons

krek1111 [17]

Cations are formed by losing electrons

Cations are basically positive ions

It's not lose or gain protons because atoms can't lose or gain protons. The protons don't move.
It's not gain electrons because electrons are negative. More electrons=more negative

If you can't remember cations are positive
CAT-ions are PAWS-itively charged

7 0

3 years ago

ASAPPP

katen-ka-za [31]

Power is the rate at which work is done (2nd option)

8 0

4 years ago

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